墨墨突然对等式很感兴趣,他正在研究a1x1+a2y2+…+anxn=B存在非负整数解的条件,他要求你编写一个程序,给定N、{an}、以及B的取值范围,求出有多少B可以使等式存在非负整数解。
maya神仙题啊,感觉自己做题难度跨度太大了qwq。
这里有一篇讲的非常好的博客https://blog.csdn.net/w4149/article/details/66476606?locationNum=3&fps=1
思路大概就是 利用取余的性质,把能够构造出来的解表示成统一的形式
发现该形式可以通过最短路更新
然后就做完了。。
#include<cstdio>
#include<algorithm>
#include<stack>
#include<queue>
#include<cmath>
#include<cstring>
#define lb(x) (x & (-x))
#define Pair pair<int, int>
#define fi first
#define se second
#define MP(x, y) make_pair(x, y)
#define LL long long
using namespace std;
const int MAXN = 1e6 + 10;
inline LL read() {
char c = getchar(); LL x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, vis[MAXN];
LL dis[MAXN], a[MAXN], Mi = 1e15, Bmin, Bmax;
void SPFA() {
queue<int> q;
memset(dis, 0xf, sizeof(dis));
dis[0] = 0; vis[0] = 1, q.push(0);
while(!q.empty()) {
int p = q.front(); q.pop(); vis[p] = 1;
for(int i = 1; i <= N; i++) {
int to = (p + a[i]) % Mi;//tag
if(dis[to] > dis[p] + a[i]) {
dis[to] = dis[p] + a[i];
if(!vis[to]) vis[to] = 1, q.push(to);
}
}
}
}
LL Query(LL x) {
LL rt = 0;
for(int i = 0; i < Mi; i++)
if(dis[i] <= x)
rt += (x - dis[i]) / Mi + 1;
return rt;
}
int main() {
N = read(); Bmin = read(); Bmax = read();
for(int i = 1; i <= N; i++) {
a[i] = read();
if(a[i] != 0) Mi = min(Mi, a[i]);
}
SPFA();
printf("%lld", Query(Bmax) - Query(Bmin - 1));
return 0;
}
/*
*/